Relativistic centripetal force

[yuiop]

Using the Lorentz transforms for translation is not justified. I gave you both the correct method based on the Lorentz transforms for rotation. If you or kev finish the computations, you are in for a big surprise.

I have finished your computations and the final result is just as messy and ugly as the equations that precede it. If your computations are correct then they have to resolve in the simplest situation to either a' = \gamma a or a' =\gamma^3 a because that is what proper acceleration is by (your) definition. It does not seem to do that but maybe I am doing it wrong. Can you demonstrate that your definition is correct and in agreement with your rotation transforms?

In your blog attachment you state:

t = \gamma \left(t ' + \frac{\omega R y '}{c^2} \right)

It seems odd in that in that expression there is no x'. Is the transformation between coordinate time and proper time really completely independent of movement in the x' direction? Somehow I doubt it because these are supposed to generalised transformations with no preferential direction.

Can you elaborate on how you arrive at the equations for dx and dy in (4)?

I have managed to locate a copy of the article you reference in your blog (Generalized Lorentz transformation for an accelerated, rotating frame of reference [J. Math. Phys. 28, 2379-2383 (1987)] Robert A. Nelson) and none of the equations in that paper match the equations in your blog. I guess that is a credit to you that you are not just copying other people's work, but since it is your work perhaps you could clarify what you are thinking. One advantage of a forum over books is supposed to be that you can ask the author what he means or to elaborate on something. All this "surprise" stuff is not very helpful. Relativity is complicated enough with plenty of opportunities for error and misunderstanding, without playing silly games.

 

[yuiop]

Let's have a look at the method you use in your generalised transformation for rotating reference frames. You start with:

\frac{dx}{dt} = \frac{dx}{dt '} \frac{dt '}{dt} = \gamma^{-1}\frac{dx}{dt '}

Fine. In the next step you obtain an equation for acceleration by:

\frac{d^2x}{dt ^2} = \frac{d^2 x}{dt '^2} \frac{dt '^2}{dt^2} = \gamma^{-2}\frac{d^2x}{dt '^2}

Rearrange:

a ' = \gamma^2 \frac{d^2x}{dt ^2}

Now by definition:

a ' = \gamma^2 \frac{d^2x}{dt ^2} = \gamma^2 \frac{dv}{dt}

which is the result obtained by pervect and me (which you say is wrong) using your method.

 

[starthaus]

I have finished your computations and the final result is just as messy and ugly as the equations that precede it.

These are not "my" computations. These are the "correct" computations for the application of the appropiate transforms. If you complete the calculations (left to you as an exercise), you will have the pleasant surprise to find the correct transformation of accelerations, which, by the way, looks very ellegant. That is, if you want to learn.

t = \gamma \left(t ' + \frac{\omega R y '}{c^2} \right)

It seems odd in that in that expression there is no x'. Is the transformation between coordinate time and proper time really completely independent of movement in the x' direction? Somehow I doubt it because these are supposed to generalised transformations with no preferential direction.

Why don't you read the reference?

Can you elaborate on how you arrive at the equations for dx and dy in (4)?

Through simple differentiation.

I have managed to locate a copy of the article you reference in your blog (Generalized Lorentz transformation for an accelerated, rotating frame of reference [J. Math. Phys. 28, 2379-2383 (1987)] Robert A. Nelson) and none of the equations in that paper match the equations in your blog. I guess that is a credit to you that you are not just copying other people's work, but since it is your work perhaps you could clarify what you are thinking.

If you have difficulty with simple math, HERE is another reference where the calculations are all done for you.

One advantage of a forum over books is supposed to be that you can ask the author what he means or to elaborate on something. All this "surprise" stuff is not very helpful. Relativity is complicated enough with plenty of opportunities for error and misunderstanding, without playing silly games.

Hey, you need to learn how to calculate for yourself, not to cherry pick from formulas derived by others.

 

[starthaus]

I read the "wrong" attachment because earlier you said:

...which is the definition of proper acceleration. This is the 4-th time I'm explaining this to you.

so naturally I assumed you said something about proper acceleration in attachment 1 but not a single equation there relates directly to acceleration or even to to a transformation between frames. Seeing nothing there,

I guess everything needs to be spelled out to you, you did not see the \frac{d}{dt}(m_0 \gamma \vec {v})=q \vec{v} X \vec{B} equation. One more time, the term \frac{d}{dt} (\gamma \vec {v}) is the proper acceleration. By defintion.

I looked in the other attachments where you do talk about acceleration, to see how you come to the conclusion that proper acceleration has a factor of gamma^3

The other attachments talk about translation motion. It is not my fault that you seem unable to tell the difference between rotation and translation.
If you were able to do that maybe you could stop ascribing me all the errors that you keep making.

 

[starthaus]

Let's have a look at the method you use in your generalised transformation for rotating reference frames. You start with:

\frac{dx}{dt} = \frac{dx}{dt '} \frac{dt '}{dt} = \gamma^{-1}\frac{dx}{dt '}

Fine.

I am glad that it met with your approval.

In the next step you obtain an equation for acceleration by:

\frac{d^2x}{dt ^2} = \frac{d^2 x}{dt '^2} \frac{dt '^2}{dt^2}

You sure about this? This is basic calculus. What you wrote is wrong, I am doing none of the stuff you are claiming I am doing. Here is what I am really doing:

a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}=\frac {d^2x}{dtdt'} \frac{dt'}{dt}

Do you see the difference? The object is to get the correct expression for a.

Rearrange:

a ' = \gamma^2 \frac{d^2x}{dt ^2}

Nope, you need to get your basic calculus straightened out. Only after that your physics will come out correct.

 

[yuiop]

You sure about this? This is basic calculus. What you wrote is wrong, I am doing none of the stuff you are claiming I am doing. Here is what I am really doing:

a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}=\frac {d^2x}{dtdt'} \frac{dt'}{dt}

Do you see the difference? The object is to get the correct expression for a.

I can extend your expression, using the methods you use, like this:

a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}= \frac {d^2x}{dtdt'} \frac{dt'}{dt} = \gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt} = \gamma^{-2} \frac {d^2x}{dt'dt'} = \gamma^{-2} \frac {d^2x}{dt'^2}

which gives:

a' = \gamma^2 \frac{d^2x}{dt ^2}

which is what I have always claimed and which you keep insisting is wrong. I think that since it obvious we are never going to agree it might be helpful if some of the pf staff gave a second opinion on the disputed items in this thread.

 

[starthaus]

I can extend your expression, using the methods you use, like this:

a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}= \frac {d^2x}{dtdt'} \frac{dt'}{dt} = \gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt} = \gamma^{-2} \frac {d^2x}{dt'dt'} = \gamma^{-2} \frac {d^2x}{dt'^2}

which gives:

a' = \gamma^2 \frac{d^2x}{dt ^2}

which is what I have always claimed and which you keep insisting is wrong. I think that since it obvious we are never going to agree it might be helpful if some of the pf staff gave a second opinion on the disputed items in this thread.

Look at the attachment I wrote for you, \frac {d^2x}{dt'^2} is not a'.
You should know better than that:

a'=\frac {d^2x'}{dt'^2}\frac {d^2x}{dt'^2} means nothing.

Your exercise is to express \frac {d^2x}{dt^2} as a function of \frac {d^2x'}{dt'^2}. I gave you all the tools to do that correctly.

 

[yuiop]

At this step:

a= \frac{d^2x}{dt dt} = \frac {d^2x}{dtdt'} \frac{dt'}{dt}

you are using the simple fact that:

\frac{1}{dt} = \frac{1}{dt'}\frac{dt'}{dt}

(Nothing wrong with that - basic algebra.) I am using exactly the same algebraic fact to complete this step:

\gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt}

Are you really saying the above equality is invalid? If so, then it you who needs to think about it.

 

[yuiop]

Look at the attachment I wrote for you, \frac {d^2x}{dt'^2} is not a'.

We are discussing centripetal acceleration which is orthogonal to the instantaneous velocity. Orthogonal distances do not length contract so dx = dx' and they are interchangeable.

\frac {d^2x}{dt'^2} is the same thing as \frac {d^2x'}{dt'^2}

You should know better than that:

a'=\frac {d^2x'}{dt'^2}

See above.

\frac {d^2x}{dt'^2} means nothing.

It means the same thing as

\frac {d^2x'}{dt'^2}

when dx' = dx, as it does in this case.

 

[starthaus]

At this step:

a= \frac{d^2x}{dt dt} = \frac {d^2x}{dtdt'} \frac{dt'}{dt}

you are using the simple fact that:

\frac{1}{dt} = \frac{1}{dt'}\frac{dt'}{dt}

(Nothing wrong with that - basic algebra.) I am using exactly the same algebraic fact to complete this step:

\gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt}

Are you really saying the above equality is invalid? If so, then it you who needs to think about it.

\frac {d^2x}{dt'^2} is physically a meaningless entity, you are mixing frames. Can you write down the correct definition for a'?

 

[starthaus]

It means the same thing as

\frac {d^2x'}{dt'^2}

when dx' = dx, as it does in this case.

:lol:

 

[yuiop]

\frac {d^2x}{dt'^2} means nothing.

which is unfortunate, because that is exactly what you derive in the final expression (6) in your blog.

 

[starthaus]

which is unfortunate, because that is exactly what you derive in the final expression (6) in your blog.

Nope. \frac{d^2x}{dt^2}. New glasses, perhaps?

 

[yuiop]

Nope. \frac{d^2x}{dt^2}. New glasses, perhaps?

You effectively derive:

\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}

in expression (6) of your attachment, although you probably don't realize that.

 

[starthaus]

You effectively derive:

\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}

I would never write such frame-mixing nonsese.

in expression (6) of your attachment, although you probably don't realize that.

You definitely need to learn how to read math :

\frac{d^2x}{dt^2} = \gamma^{-2}(\frac{d^2x'}{dt'^2}+...

Can you see the pairing x,t in the LHS and the pairing x',t' in the RHS?
If you were less obsessed with finding errors where there are none, maybe you would be more able to learn.

 

[yuiop]

You effectively derive:

\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}

I would never write such frame-mixing nonsese.

You wrote such frame-mixing nonsense at step (5) when you derive the velocity transformation as:

\frac{dx}{dt} = \gamma^{-1}\frac{dx}{dt'}

 

[starthaus]

You wrote such frame-mixing nonsense at step (5) when you derive the velocity transformation as:

\frac{dx}{dt} = \gamma^{-1}\frac{dx}{dt'}

You definitely need a new pair of glasses :


\frac{dx}{dt} = \gamma^{-1}(\frac{dx'}{dt'}+...

Are you getting that desperate to prove me wrong that you can't even follow simple arithmetic anymore?

 

[Jorrie]

You wrote such frame-mixing nonsense at step (5) when you derive the velocity transformation as:

\frac{dx}{dt} = \gamma^{-1}\frac{dx}{dt'}

I have a feeling that the reason behind this 'brutal' argument is a miscommunication. Starthaus's full equations treat the general solution of rotating to Cartesian coordinate transformations. In this case, saying that dx' and dx are equivalent is not valid.

Kev and myself were mainly interested in one point only, say when omega=0, y=0 and dx' and dx are momentarily equivalent. If we have the magnitude of the proper acceleration at that point, we have it for all time (e.g., the reading of the mass on the scale against the inside wall of the rotating cylinder will remain constant). That way we simplify the route to a simple result.

 

[starthaus]

I have a feeling that the reason behind this 'brutal' argument is a miscommunication. Starthaus's full equations treat the general solution of rotating to Cartesian coordinate transformations. In this case, saying that dx' and dx are equivalent is not valid.

Good, you are making the effort to understand rather than find fault, like kev.

Kev and myself were mainly interested in one point only, say when omega=0,

\omega cannot be ever zero, the frame is totating.

dx' and dx are momentarily equivalent.

This never happens. Look at the expression of dx, it is a function of (dx',dy',t')

There is something that you can do , though. You can consider the point (0,0) in the frame S'. When you do that you will obtain the correct expression for the transformation between frames but you need to do the math correctly, without attempting ugly hacks like setting dx=dx'.

 

[Jorrie]

\omega cannot be ever zero, the frame is totating.

Sorry, I actually meant when \theta = 0, where \omega=d\theta/dt

It is perhaps time that you clear up your attachment - the first line of your equations [5] of "Uniformly Rotating Frames" does use the contentious 'mixing of frames' (dx/dt').

It would also be immensely helpful to all around here (especially me as an engineer, no mathematician) if you would correct the typo and complete equations [6]...

 

[yuiop]

\omega cannot be ever zero, the frame is totating.

Jorrie obviously meant that when the angle (theta or phi or whatever you choose to call it) between the x axes of the two frames is zero and by the definition given in this link that you gave http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf this occurs when t' =0.

dx' and dx are momentarily equivalent.

This never happens. Look at the expression of dx, it is a function of (dx',dy',t')

It does happen. At time t' = 0, the axes are aligned and x = x' = 0 and y = y' =0.
By expression (4) of your attachment, when t' = 0, all the terms containing sin(y\omega t') vanish and the term containing y' vanishes too, leaving

dx = dx'cos(0) = dx'.

Although that seems trivial, the same is not true for dy at time t'=0.
The dy equation in expression (4) becomes:

dy = \gamma dy' +R\gamma\omega dt'

 

[starthaus]

Sorry, I actually meant when \theta = 0, where \omega=d\theta/dt

It is perhaps time that you clear up your attachment - the first line of your equations [5] of "Uniformly Rotating Frames" does use the contentious 'mixing of frames' (dx/dt').

No, it doesn't. It contains the standard "chain rule" for calculating derivatives.

\frac {dx}{dt}=\frac{dx}{dt'} \frac{dt'}{dt}

This is standard math, not the type of frame-mixing that kev does.

It would also be immensely helpful to all around here (especially me as an engineer, no mathematician) if you would correct the typo and complete equations [6]...

I am quite sure that you and kev can complete expressions (6). I left them unfiinished on purpose, to lead you to the correct results and to teach you the correct way of solving this problem. If I give you the result ready made, kev will continue to find imaginary errors. Much better if I guide you to finding it.

 

[starthaus]

Jorrie obviously meant that when the angle (theta or phi or whatever you choose to call it) between the x axes of the two frames is zero and by the definition given in this link that you gave http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf this occurs when t' =0.


It does happen. At time t' = 0, the axes are aligned and x = x' = 0 and y = y' =0.
By expression (4) of your attachment, when t' = 0, all the terms containing sin(y\omega t') vanish and the term containing y' vanishes too, leaving

dx = dx'cos(0) = dx'.

True, it happens in ONE PARTICULAR case. It does NOT happen in ANY other INFINITE number of cases. What does this tell you?

Although that seems trivial, the same is not true for dy at time t'=0.
The dy equation in expression (4) becomes:

dy = \gamma dy' +R\gamma\omega dt'

Yep, it NEVER happens. What does this tell you? Do you still claim that you can use the expressions derived from Lorentz transforms for translational motion?

 

[Dale]

I hesitate to get involved in a good sparring match, but I have an old Mathematica notebook on relativistic uniform circular motion using four-vectors. I could post the results here if that would be helpful in any way.

 

[starthaus]

I hesitate to get involved in a good sparring match, but I have an old Mathematica notebook on relativistic uniform circular motion using four-vectors. I could post the results here if that would be helpful in any way.

I think that it is probably best to let kev work out the answers on his own. Only this way will he learn, giving away the answers is not the best way to teach somebody. He has been given all the tools, it is now up to him to complete the derivation.

 

[yuiop]

I hesitate to get involved in a good sparring match, but I have an old Mathematica notebook on relativistic uniform circular motion using four-vectors. I could post the results here if that would be helpful in any way.

I would be very interested in your contribution. I think this thread is well overdue some input from the more knowledgeable members of this forum.

 

[yuiop]

I think that it is probably best to let kev work out the answers on his own. Only this way will he learn, giving away the answers is not the best way to teach somebody. He has been given all the tools, it is now up to him to complete the derivation.

This thread is not just about "teach kev a lesson". It's primary objective is to discuss the issues involved in relativistic centripetal force and acceleration in a way that might be useful for members of this forum. If that means I, you, Jorrie or anyone else learns something along the way, then all well and good. It is not just about your ego. A think a worthy secondary objective would be debug the document you presented in your blog, so that it might become a useful reference document for members of this forum. At this moment in time it far from status.

 

[yuiop]

dx = dx'cos(0) = dx'.

True, it happens in ONE PARTICULAR case. It does NOT happen in ANY other INFINITE number of cases. What does this tell you?

That tells me that the x and y components are changing over time, but because of the circular symmetry and because the rotation is uniform and constant, the angular velocity, centripetal force, centripetal acceleration and gamma factor all remain constant over time and for any any angle theta and those are the physical quantities jorrie and I are interested in. Your equations seem unable to determine those quantities.

Do you still claim that you can use the expressions derived from Lorentz transforms for translational motion?

Yes I do when we talk about MCIF's and proper acceleration as measured by an accelerometer. Your equations are about acceleration in terms of spatial displacement where a glass paperweight on your table has zero acceleration because its location is not changing over time, while an accelerometer will show the paperweight is accelerating despite the fact it appears to be stationary in your accelerating reference frame.


The reference paper that you lifted the initial eqations from http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf explicitly disagrees with your last statement when they say at (19):

In particular, at t′ = 0 these transformations ... coincide with the ordinary Lorentz boost at t′ = 0 for the velocity in the y-direction.

and later on when they say in Section 6:

Therefore, for a small range of values of t′, the transformations (6)-(7) can be approximated by the ordinary Lorentz boosts (see (19)). From this fact we conclude that if a moving rigid body is short enough, then its relativistic contraction in the direction of the instantaneous velocity, as seen from S, is simply given by L(t) = L′/(t), i.e., it depends only on the instantaneous velocity, not on its acceleration and rotation. (“Short enough” means that L′ ≪ c2/a′ k, where a′k is the component of the proper acceleration parallel to the direction of the velocity [11]).
By a similar argument we may conclude that an arbitrarily accelerated and rotating observer sees equal lengths of other differently moving objects as an inertial observer whose instantaneous position and velocity are equal to that of the arbitrarily accelerated and rotating observer.

 

[starthaus]

This thread is not just about "teach kev a lesson". It's primary objective is to discuss the issues involved in relativistic centripetal force and acceleration in a way that might be useful for members of this forum. If that means I, you, Jorrie or anyone else learns something along the way, then all well and good. It is not just about your ego.

It's not about my ego. It is about you stopping trying to find errors where they don't exist and starting to do a couple of differentiations such that you find the answer for yourself. (the first differentiation is already done, you only need to complete the second order).

A think a worthy secondary objective would be debug the document you presented in your blog,

There is no "debug". There is rolling up your sleeves and calculating how acceleration transforms under rotation using the appropiate Lorentz transforms.
I will give you a hint: the method that you need to apply is identical to the one one uses to derive acceleration from the Lorentz transforms for translation. Except that you need to use the Lorentz transforms for rotation that have been provided for you. Rather than arguing for days about imaginary errors you could have obtained the answer by now.

 

[yuiop]

Your exercise is to express \frac {d^2x}{dt^2} as a function of \frac {d^2x'}{dt'^2}. I gave you all the tools to do that correctly.

You have not obtained \frac {d^2x'}{dt'^2} in your document. All you have done is taken the derivative of x with respect to time twice and then transformed the result by converting dt to dt' using dt = gamma(dt'), What you have failed to do, is convert dx to dx' using dx = f(dx') where f is the long function inplicit in expression (4) of your document.

I am quite sure that you and kev can complete expressions (6). I left them unfiinished on purpose, to lead you to the correct results and to teach you the correct way of solving this problem. If I give you the result ready made, kev will continue to find imaginary errors. Much better if I guide you to finding it.

It is easy to complete expression (6) because it is obvious from the terms you have completed and from the method you use in obtaining expression (5), that all you are doing is dividing each term by \gamma dt'. The completed expressions are (as far as I can tell):

\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - \frac{d}{dt'}R\gamma\omega\sin(\gamma\omega t')\right)

\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') + \frac{d}{dt'}R\gamma\omega\cos(\gamma\omega t')\right)

If that is not what you intended, I have done the hard work of formatting the tex and all you have to do is edit it.

Note that for constant uniform rotation, the last term of each expression containg R probably vanishes, because none of the quantities R,\gamma,\omega are changing over time.

Also note that neither jorrie or myself claim to be expert mathematicians and it would be helpful to both of us if you would break down how you get from (1) to (4). After all, you claim your goal is educate me.